Solve [(m+2)+(m-3)]^2-[(m+2)-(m-3)]^2=24

$$[ ( m + 2 ) + ( m - 3 ) ] ^ { 2 } - [ ( m + 2 ) - ( m - 3 ) ] ^ { 2 } = 24$$

Solve for m

$$m=-3$$

$$m=4$$

Steps Using Factoring By Grouping

Combine m and m to get 2m.

$$\left(2m+2-3\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Subtract 3 from 2 to get -1.

$$\left(2m-1\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-1\right)^{2}.

$$4m^{2}-4m+1-\left(m+2-\left(m-3\right)\right)^{2}=24$$

To find the opposite of m-3, find the opposite of each term.

$$4m^{2}-4m+1-\left(m+2-m+3\right)^{2}=24$$

Combine m and -m to get 0.

$$4m^{2}-4m+1-\left(2+3\right)^{2}=24$$

Add 2 and 3 to get 5.

$$4m^{2}-4m+1-5^{2}=24$$

Calculate 5 to the power of 2 and get 25.

$$4m^{2}-4m+1-25=24$$

Subtract 25 from 1 to get -24.

$$4m^{2}-4m-24=24$$

Subtract 24 from both sides.

$$4m^{2}-4m-24-24=0$$

Subtract 24 from -24 to get -48.

$$4m^{2}-4m-48=0$$

Divide both sides by 4.

$$m^{2}-m-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite m^{2}-m-12 as \left(m^{2}-4m\right)+\left(3m-12\right).

$$\left(m^{2}-4m\right)+\left(3m-12\right)$$

Factor out m in the first and 3 in the second group.

$$m\left(m-4\right)+3\left(m-4\right)$$

Factor out common term m-4 by using distributive property.

$$\left(m-4\right)\left(m+3\right)$$

To find equation solutions, solve m-4=0 and m+3=0.

$$m=4 m=-3$$

Steps Using the Quadratic Formula

Combine m and m to get 2m.

$$\left(2m+2-3\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Subtract 3 from 2 to get -1.

$$\left(2m-1\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-1\right)^{2}.

$$4m^{2}-4m+1-\left(m+2-\left(m-3\right)\right)^{2}=24$$

To find the opposite of m-3, find the opposite of each term.

$$4m^{2}-4m+1-\left(m+2-m+3\right)^{2}=24$$

Combine m and -m to get 0.

$$4m^{2}-4m+1-\left(2+3\right)^{2}=24$$

Add 2 and 3 to get 5.

$$4m^{2}-4m+1-5^{2}=24$$

Calculate 5 to the power of 2 and get 25.

$$4m^{2}-4m+1-25=24$$

Subtract 25 from 1 to get -24.

$$4m^{2}-4m-24=24$$

Subtract 24 from both sides.

$$4m^{2}-4m-24-24=0$$

Subtract 24 from -24 to get -48.

$$4m^{2}-4m-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-48\right)}}{2\times 4}$$

Square -4.

$$m=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-48\right)}}{2\times 4}$$

Multiply -4 times 4.

$$m=\frac{-\left(-4\right)±\sqrt{16-16\left(-48\right)}}{2\times 4}$$

Multiply -16 times -48.

$$m=\frac{-\left(-4\right)±\sqrt{16+768}}{2\times 4}$$

Add 16 to 768.

$$m=\frac{-\left(-4\right)±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$m=\frac{-\left(-4\right)±28}{2\times 4}$$

The opposite of -4 is 4.

$$m=\frac{4±28}{2\times 4}$$

Multiply 2 times 4.

$$m=\frac{4±28}{8}$$

Now solve the equation m=\frac{4±28}{8} when ± is plus. Add 4 to 28.

$$m=\frac{32}{8}$$

Divide 32 by 8.

$$m=4$$

Now solve the equation m=\frac{4±28}{8} when ± is minus. Subtract 28 from 4.

$$m=-\frac{24}{8}$$

Divide -24 by 8.

$$m=-3$$

The equation is now solved.

$$m=4 m=-3$$

Steps for Completing the Square

Combine m and m to get 2m.

$$\left(2m+2-3\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Subtract 3 from 2 to get -1.

$$\left(2m-1\right)^{2}-\left(m+2-\left(m-3\right)\right)^{2}=24$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2m-1\right)^{2}.

$$4m^{2}-4m+1-\left(m+2-\left(m-3\right)\right)^{2}=24$$

To find the opposite of m-3, find the opposite of each term.

$$4m^{2}-4m+1-\left(m+2-m+3\right)^{2}=24$$

Combine m and -m to get 0.

$$4m^{2}-4m+1-\left(2+3\right)^{2}=24$$

Add 2 and 3 to get 5.

$$4m^{2}-4m+1-5^{2}=24$$

Calculate 5 to the power of 2 and get 25.

$$4m^{2}-4m+1-25=24$$

Subtract 25 from 1 to get -24.

$$4m^{2}-4m-24=24$$

Add 24 to both sides.

$$4m^{2}-4m=24+24$$

Add 24 and 24 to get 48.

$$4m^{2}-4m=48$$

Divide both sides by 4.

$$\frac{4m^{2}-4m}{4}=\frac{48}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$m^{2}+\left(-\frac{4}{4}\right)m=\frac{48}{4}$$

Divide -4 by 4.

$$m^{2}-m=\frac{48}{4}$$

Divide 48 by 4.

$$m^{2}-m=12$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-m+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-m+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$m^{2}-m+\frac{1}{4}=\frac{49}{4}$$

Factor m^{2}-m+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$m-\frac{1}{2}=\frac{7}{2} m-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$m=4 m=-3$$